# Given I n </msub> = <msubsup> &#x222B;<!-- ∫ --> <mrow class="MJX-TeXAtom

Given
${I}_{n}={\int }_{-1}^{1}\left(1-{x}^{2}{\right)}^{n}\mathrm{cos}\left(\theta x\right)dx,$
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Tanner Hamilton
Looks like it will be easier to prove it using strong induction on n
Show the base case holds. Assume the induction hypothesis for $k\le n$, ie,
$\begin{array}{}\text{(}†\text{)}& \mathrm{\forall }k\le n:\phantom{\rule{1em}{0ex}}{\theta }^{2k+1}{I}_{k}=k!\left({P}_{k}\left(\theta \right)\mathrm{sin}\left(\theta \right)+{Q}_{k}\left(\theta \right)\mathrm{cos}\left(\theta \right)\right)\end{array}$
Now, show the inductive step for $n+1$, ie,
$\begin{array}{rl}{\theta }^{2\left(n+1\right)+1}{I}_{n+1}& ={\theta }^{2n+1}{\theta }^{2}{I}_{n+1}\\ & \stackrel{\left(1\right)}{=}{\theta }^{2n+1}\left(\left(2n+2\right)\left(2n+1\right){I}_{n}-\left(2n+2\right)\left(2n\right){I}_{n-1}\right)\\ & =\left(2n+2\right)\left(2n+1\right)\left[{\theta }^{2n+1}{I}_{n}\right]-\left(2n+2\right)\left(2n\right){\theta }^{2}\left[{\theta }^{2\left(n-1\right)+1}{I}_{n-1}\right]\\ & \stackrel{\left(†\right)}{=}\left(2n+2\right)\left(2n+1\right)n!\left({P}_{n}\left(\theta \right)\mathrm{sin}\theta +{Q}_{n}\left(\theta \right)\mathrm{cos}\theta \right)-{\theta }^{2}\left(2n+2\right)\left(2n\right)\left(\left(n-1\right)!{P}_{n-1}\left(\theta \right)\mathrm{sin}\theta +{Q}_{n-1}\mathrm{cos}\theta \right)\\ & =2\left(2n+1\right)\left(n+1\right)!\left({P}_{n}\left(\theta \right)\mathrm{sin}\theta +{Q}_{n}\left(\theta \right)\mathrm{cos}\theta \right)-4\left(n+1\right)!\left({\theta }^{2}{P}_{n-1}\left(\theta \right)\mathrm{sin}\theta +{\theta }^{2}{Q}_{n-1}\mathrm{cos}\theta \right)\\ & =\left(n+1\right)!\left({P}_{n+1}\left(\theta \right)\mathrm{sin}\theta +{Q}_{k+1}\left(\theta \right)\mathrm{cos}\theta \right)\end{array}$
where ${P}_{n+1}\left(\theta \right):=2\left(2n+1\right){P}_{n}\left(\theta \right)-4{\theta }^{2}{P}_{n-1}\left(\theta \right)$ and ${Q}_{n+1}\left(\theta \right):=2\left(2n+1\right){Q}_{n}\left(\theta \right)-4{\theta }^{2}{Q}_{n-1}\left(\theta \right)$ and we can finish noting that ${P}_{n+1}$ and ${Q}_{n+1}$ are both polynomials with integer coefficients and degree at most $2n+2=2\left(n+1\right)$